The first compound I am going to assign HNMR peaks to is Piperonal.
as i assigned number 1-5, i have identified 5 different types of hydrogens that are going to show the five different peaks on the HNMR spectrum. Hydrogens labled as number are equivilent so there are no coupling and are going to show a singlet but with the height of 2H. (S,2H). Hydrogen labled as 2 is going to show a doublet (d,1H) because of spin-spin splitting with Hydrogen 3 on the same aeromatic ring. Hydrogen 2 and 3 are adjacent to each other which makes them have the spin-spin splitting. the J constant for this Hydrogen is about 8Hz which we can calculate by the difference of the ranges in PPM an multiplying by 300 (DFRQ# given). and after the calcualtion i have found it to be about 7.2 which is close to 8 is valid for out argument. Hydrogen 3 show two spin-spin splitting because it it adjacent to Hydrogen 2 and meta relation to Hydrogen 5 on the aeromatic ring. so Hydrogen 3 has (d,d,1H) two doublets next to each other. Hydrogen 4 is an aldehyde and had no coupling effect from the aeromatic ring and therefore shows only a singlet (S,1H). also we know it is going to lie between the ranges of 9-10 because it's an aldehyde. As i mentioned before Hydrogen 5 is experiencing meta coupling effect because of Hydrogen 3 on the aeromatic ring so it is going to show a doublet (d,1H). Hydrogens 2,3,and 5 all lie between the ranges of about 7-8. Also Hydrogen 5 is going to have a J constant of about 2Hz. After the integration we see that our assumtions are correct and the length of the integration lines corelates to the numbers of hydrogen.
Grade 7/10
The dd of H3 has 2 J constants and is neither is 0.18 Hz.
No link to origin of spectrum.
Too many spelling errors.
as i assigned number 1-5, i have identified 5 different types of hydrogens that are going to show the five different peaks on the HNMR spectrum. Hydrogens labled as number are equivilent so there are no coupling and are going to show a singlet but with the height of 2H. (S,2H). Hydrogen labled as 2 is going to show a doublet (d,1H) because of spin-spin splitting with Hydrogen 3 on the same aeromatic ring. Hydrogen 2 and 3 are adjacent to each other which makes them have the spin-spin splitting. the J constant for this Hydrogen is about 8Hz which we can calculate by the difference of the ranges in PPM an multiplying by 300 (DFRQ# given). and after the calcualtion i have found it to be about 7.2 which is close to 8 is valid for out argument. Hydrogen 3 show two spin-spin splitting because it it adjacent to Hydrogen 2 and meta relation to Hydrogen 5 on the aeromatic ring. so Hydrogen 3 has (d,d,1H) two doublets next to each other. Hydrogen 4 is an aldehyde and had no coupling effect from the aeromatic ring and therefore shows only a singlet (S,1H). also we know it is going to lie between the ranges of 9-10 because it's an aldehyde. As i mentioned before Hydrogen 5 is experiencing meta coupling effect because of Hydrogen 3 on the aeromatic ring so it is going to show a doublet (d,1H). Hydrogens 2,3,and 5 all lie between the ranges of about 7-8. Also Hydrogen 5 is going to have a J constant of about 2Hz. After the integration we see that our assumtions are correct and the length of the integration lines corelates to the numbers of hydrogen.
Grade 7/10
The dd of H3 has 2 J constants and is neither is 0.18 Hz.No link to origin of spectrum.
Too many spelling errors.